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1 Internal energy \((U)\)
2 Thermodynamic temperature \((T)\)
Q: What does A-Level Physics: 13) Thermodynamic Systems Guide cover? A: Internal energy, pΔV work, the Zeroth & First Laws, plus specific heat and latent heat - this guide unpacks Topic 13 of the 2026 H2 Physics syllabus.
TL;DR Thermodynamics is the energy balance sheet of the universe. Nail the distinctions between U, Q, W and you will unlock marks in gas laws, kinetic theory and every practical calorimetry task.
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If you only have...
Focus on...
1 second
Track internal energy, heating, and work separately
10 seconds
(U), (Q), (W), and the First Law
100 seconds
Internal energy and the first mini-drill
10 minutes
p-V work, heat capacity, latent heat, and cycles
Concrete example: how to use this page
For a gas in a cylinder, ask three questions: did heat enter, did the gas do work, and did internal energy change? Answer those before substituting into the First Law.
Stay synced with the adjacent thermal-physics chapters via the H2 Physics notes hub; it bundles Temperature & Ideal Gases plus exam drills referenced here.
1 Internal energy (U)
The macroscopic state of a system fixes its internal energy: the sum of
random
microscopic kinetic (Ek) and potential (Ep) energies. No single thermometer can reveal U directly - you must track
how
energy enters or leaves.
1.1 Parent takeaway
A student who writes “heat stored in the gas” is forfeiting method marks. Heat is a transfer, not a store.
1.2 Mini-drill
State whether each contributes to U:
Scenario
Contributes to U?
Translational motion of the whole cylinder
No
Vibrations between gas molecules
Yes
2 Thermodynamic temperature (T)
Absolute temperature in kelvin is directly proportional to the mean kinetic energy per particle. Doubling T doubles ⟨Ek⟩. Always quote exam answers to 2-3 s.f. unless otherwise stated.
3 Heating (Q) vs work (W)
Symbol
Process
Positive when...
Typical formula
Q
Energy transfer by heating
Energy enters system
Q=mcΔT
W
Mechanical work
Done on system
W=−pΔV (constant p; ΔV=Vf−Vi
Sign hack: In SEAB mark schemes, work done by the gas is negative.
4 Zeroth law - the thermometer principle
If A is in thermal equilibrium with B, and B with C, then A is in equilibrium with C. This justifies using a third body (a thermometer) to compare temperatures.
5 First law - the energy ledger
ΔU=Q+W.
Positive Q: heating adds energy.
Positive W: surroundings compress system.
For an ideal gas in free expansion, Q=0 and W=0, hence ΔU=0: temperature unchanged.
6 Specific heat capacity (c)
Defined as the heat required per unit mass per kelvin rise:
Q=mcΔT.
6.1 Quick-check
Water has c≈4.18kJ⋅kg−1⋅K−1 - roughly 11x that of copper (≈0.39kJ⋅kg−1⋅K−1). That is why water is a coolant.
7 Specific latent heat (L)
Energy per unit mass for a phase change at constant T:
Q=mL.
Common values:\
Substance
Lf (fusion) / kJ⋅kg−1
Lv (vaporisation) / kJ⋅kg−1
Water
334
2260
8 WA & Paper 4 timing rules
Use syllabus pacing as a guide: Paper 2/3 average ~1.6 min/mark; Paper 4 ~3 min/mark.
Copy units first when using data-book tables.
Show full working; SEAB awards method marks even if arithmetic slips.
Need structured practice on Thermodynamic Systems? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.
Comprehensive revision pack
9478 Section IV, Topic 13 Syllabus outcomes
Candidates should be able to:
(a) show an understanding that the macroscopic state of a system determines the internal energy of the system, and that internal energy can be expressed as the sum of a random distribution of microscopic kinetic and potential energies associated with the particles of the system.
(b) show an understanding that the thermodynamic temperature of a system is (directly) proportional to the mean microscopic kinetic energy of particles.
(c) show an understanding that when two systems are placed in thermal contact, energy is transferred (by heating) from the system at higher temperature to the system at lower temperature, until they reach the same temperature and achieve thermal equilibrium (i.e. no net energy transfer).
(d) show an understanding of the difference between the work done by a gas and the work done on a gas, and calculate the work done by a gas in expanding against a constant external pressure: W=pΔV.
(e) recall and apply the zeroth law of thermodynamics that if two systems are both in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other.
(f) recall and apply the first law of thermodynamics, ΔU=Q+W, that the increase in internal energy of a system is equal to the sum of the energy transferred to the system by heating and the work done on the system.
(g) define and use the concepts of specific heat capacity and specific latent heat.
Concept map (in words)
Identify the process path (constant T, V, p or adiabatic). Track energy transfers: heating (Q) and work (W) alter internal energy (U). Use c and L for temperature changes and phase changes respectively. p-V diagrams visualise work as area under curve.
Key relations
Quantity/Process
Expression / reminder
First law
ΔU=Q+W (W positive when done on system)
Work at constant pressure
W=pΔV (positive if compression)
Specific heat
Q=mcΔT
Specific latent heat
Q=mL
Isothermal ideal gas
pV=constant, ΔU=0
Isochoric process
ΔV=0⇒W=0
Adiabatic (ideal gas)
Q=0⇒ΔU=W
Internal energy of ideal gas
U=23nRT (monatomic)
Derivations & reasoning to master
First law sign convention: derive for compression/expansion, noting SEAB's “work done by gas is negative”.
Work on a p-V diagram: show W=∫pdV and interpret the area under the curve.
Heating vs temperature change: explain why adding latent heat occurs without temperature change.
Adiabatic vs isothermal: compare outcomes for ΔU to emphasise physical differences.
Worked example 1 - heating and work
An ideal gas is compressed slowly from 4.0L to 2.0L at constant pressure 150kPa while 1.6kJ of heat is removed. Find the change in internal energy.
Method: W=−pΔV (constant p); apply ΔU=Q+W. Discuss the sign of ΔU (gas cooled).
ΔV=2.0−4.0=−2.0L⇒W=−pΔV=−150(−2.0)=+300J=0.30kJ.
Heat is removed so Q=−1.6kJ. Hence
ΔU=Q+W=(−1.6+0.30)kJ=−1.3kJ.
Worked example 2 - latent heat application
0.45kg of ice at −12∘C is heated to steam at 110∘C. Calculate the total energy required. Use cice, cwater, csteam, Lf, Lv.
Approach: break into stages: warm ice → melt → warm water → vaporise → superheat steam. Sum each Q.
Using typical data-book values cice≈2.1kJ⋅kg−1⋅K−1, cwater≈4.18kJ⋅kg−1⋅K−1, csteam≈2.0kJ⋅kg−1⋅K−1, Lf=334kJ⋅kg−1, Lv=2260kJ⋅kg−1:
Parents: schedule a 1-hour Thermodynamic Systems clinic two weeks before WA 2.
Students: print the First Law equation and stick it on your water bottle - then apply it to every past-year Qn you meet.
Last updated 14 Jul 2025. Next review: upon release of the 2027 draft syllabus.
